1.

A particle of mass m kg and charge q coulomb is accelerated through V volt then the de-Broglie wavelength associated with it,in meter is lambda=…..m.

Answer»

<P>`lambda=(h)/(mv)`
`lambda=(h)/(sqrt(2mqV))`
`lambda=(h)/(sqrt(mqV))`
`lambda=(hq)/(sqrt(2mV))`

Solution :de-Broglie wavelength `lambda=(h)/(p)`….(1)
`(p^(2))/(2M)=qVimpliesp=sqrt(2Vqm)`……..(2)
PUTTING vallue of EQUATION (2) in equation (1),
`therefore lambda=(h)/(sqrt(2Vqm))`


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